# How do I sum the series #sum_(n=1)^oon(3/4)^n#?

The answer is: 12.

First of all: "How much is the sum of a geometric series?".

Let's list all the terms of the series:

or, better:

The first sum is:

The second sum is:

The third sum is:

The fourth sum is:

So:

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To sum the series ( \sum_{n=1}^{\infty} n \left(\frac{3}{4}\right)^n ), you can use the formula for the sum of an infinite geometric series:

[ S = \frac{a}{1 - r} ]

where ( a ) is the first term of the series and ( r ) is the common ratio.

In this series, the first term (( a )) is ( \frac{3}{4} ), and the common ratio (( r )) is also ( \frac{3}{4} ) because each term is obtained by multiplying the previous term by ( \frac{3}{4} ).

Thus, plugging these values into the formula:

[ S = \frac{\frac{3}{4}}{1 - \frac{3}{4}} ]

[ S = \frac{\frac{3}{4}}{\frac{1}{4}} ]

[ S = 3 ]

So, the sum of the series ( \sum_{n=1}^{\infty} n \left(\frac{3}{4}\right)^n ) is 3.

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